Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{k + 9}{2k + 12} \div \dfrac{-6k - 54}{k^2 + 14k + 48} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{k + 9}{2k + 12} \times \dfrac{k^2 + 14k + 48}{-6k - 54} $ First factor the quadratic. $n = \dfrac{k + 9}{2k + 12} \times \dfrac{(k + 6)(k + 8)}{-6k - 54} $ Then factor out any other terms. $n = \dfrac{k + 9}{2(k + 6)} \times \dfrac{(k + 6)(k + 8)}{-6(k + 9)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (k + 9) \times (k + 6)(k + 8) } { 2(k + 6) \times -6(k + 9) } $ $n = \dfrac{ (k + 9)(k + 6)(k + 8)}{ -12(k + 6)(k + 9)} $ Notice that $(k + 9)$ and $(k + 6)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(k + 9)}(k + 6)(k + 8)}{ -12\cancel{(k + 6)}(k + 9)} $ We are dividing by $k + 6$ , so $k + 6 \neq 0$ Therefore, $k \neq -6$ $n = \dfrac{ \cancel{(k + 9)}\cancel{(k + 6)}(k + 8)}{ -12\cancel{(k + 6)}\cancel{(k + 9)}} $ We are dividing by $k + 9$ , so $k + 9 \neq 0$ Therefore, $k \neq -9$ $n = \dfrac{k + 8}{-12} $ $n = \dfrac{-(k + 8)}{12} ; \space k \neq -6 ; \space k \neq -9 $